Question

Determine o rotacional do campo vetorial V(x,y,z) = (xyz, 0, −x²y).

Answer 1

✅ Após resolver os cálculos, concluímos que o rotacional do referido campo vetorial é:

          $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\textrm{rot}\:\vec{V} = -x^{2}\,\vec{i} + 3xy\,\vec{j} - xz\,\vec{k}\:\:\:}}\end{gathered} }$

   

Seja a função:

              $\Large\displaystyle\begin{gathered} V(x, y, z) = (xyz,\,0,\,-x^{2}y)\end{gathered}$

Organizando o campo vetorial, temos:

        $\Large\displaystyle\begin{gathered} \vec{V}(x, y, z) = (xyz)\vec{i} + (0)\vec{j} + (-x^{2}y)\vec{k}\end{gathered}$

Sendo V um campo vetorial em R³, podemos dizer que o rotacional de V - denotado por "rot F" - é o produto vetorial entre o operador diferencial e V, isto é:

    $\Large\displaystyle\begin{gathered} \textrm{rot}\:\vec{V} = \nabla\wedge\vec{V}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = \bigg(\frac{\partial}{\partial x},\,\frac{\partial}{\partial y},\,\frac{\partial}{\partial z}\bigg) \wedge(X_{V}\vec{i},\,Y_{V}\vec{j},\,Z_{V}\vec{k})\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\X_{V} & Y_{V} & Z_{V}\end{vmatrix}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = \begin{vmatrix}\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\Y_{V} & Z_{V}\end{vmatrix}\vec{i} - \begin{vmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\X_{V} & Z_{V}\end{vmatrix}\vec{j} + \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y}\\X_{V} & Y_{V}\end{vmatrix}\vec{k}\end{gathered}$

                $\large\displaystyle\text{ \begin{gathered} = \left(\frac{\partial Z_{V}}{\partial y} - \frac{\partial Y_{V}}{\partial z}\right)\vec{i} - \left(\frac{\partial Z_{V}}{\partial x} - \frac{\partial X_{V}}{\partial z}\right)\vec{j} + \left(\frac{\partial Y_{V}}{\partial x} - \frac{\partial X_{V}}{\partial y}\right)\vec{k}\end{gathered} }$

                   $\Large\displaystyle\begin{gathered} = (-x^{2} - 0)\vec{i} - (-2xy - xy)\vec{j} + (0 - xz)\vec{k}\end{gathered}$

                   $\Large\displaystyle\begin{gathered} = -x^{2}\,\vec{i} + 3xy\,\vec{j} - xz\,\vec{k}\end{gathered}$      

✅ Portanto, a resposta é:

         $\Large\displaystyle\begin{gathered} \textrm{rot}\:\vec{V} = -x^{2}\,\vec{i} + 3xy\,\vec{j} - xz\,\vec{k}\end{gathered}$

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